#### Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 25

$k=6$

Hint:

$f\left ( x \right )$ must be defined. The limit of the  $f\left ( x \right )$ approaches the value $k$ must exist.

Given:

# $f(x)=\left\{\begin{array}{c} \frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2} \\\\ 3, \text { if } x=\frac{\pi}{2} \end{array}\right.$

Solution:

$f(x)=\left\{\begin{array}{c} \frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2} \\\\ 3, \text { if } x=\frac{\pi}{2} \end{array}\right.$

If $f\left ( x \right )$ is continuous at $x=\frac{\pi}{2}$ , then

\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\\\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=3 \end{aligned}                                                                       ......(i)

Putting $x=\frac{\pi}{2}-h$

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}$

From (i)

\begin{aligned} &\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}=3 \\\\ &\lim _{h \rightarrow 0} \frac{k \sinh }{2 h}=3 \\\\ &\lim _{h \rightarrow 0} \frac{k \sinh }{h}=6 \end{aligned}

$k \lim _{h \rightarrow 0} \frac{\sinh }{h}=6$                                                                                     $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$k\times 1= 6$

$k=6$, $f\left ( x \right )$  is continuous at $x=\frac{\pi}{2}$ .